-1

0

1

2

3

4

5

6

0

10

20

30

40

50

TIME (µs)

VOUT

Va

Vc

CC

2

3

A2

1

2

3

V

(R

R )

V

R

(R

R )

=

+

P

P

CC

2

A1

2

1

3

V

R

V

R

R

R

´

=

+

P

21

LPV7215

www.ti.com

SNOSAI6J – SEPTEMBER 2005 – REVISED AUGUST 2016

Product Folder Links: LPV7215

Submit Documentation Feedback

Copyright © 2005–2016, Texas Instruments Incorporated

Typical Applications (continued)

Consider the output of Figure 38 to be high to analyze the circuit. That implies that the inverted input (VC) is

lower than the noninverting input (VA). This causes the C1 to be charged through R4, and the voltage VC

increases until it is equal to the noninverting input. The value of VA at this point is in Equation 8.

(8)

If R1 = R2 = R3 then VA1 = 2 VCC/3

At this point the comparator switches pulling down the output to the negative rail. The value of VA at this point, as

shown in Equation 9:

(9)

If R1 = R2 = R3 then VA2 = VCC/3

The capacitor C1 now discharges through R4, and the voltage VC decreases until it is equal to VA2, at which point

the comparator switches again, bringing it back to the initial stage. The time period is equal to twice the time it

takes to discharge C1 from 2 VCC/3 to VCC/3, which is given by R4C1 × ln2. Hence the formula for the frequency is

given by Equation 10:

F = 1/(2 × R4 × C1 × ln2)

(10)

8.2.1.3 Application Curves

Figure 39 shows the simulated results of an oscillator using the following values:

1. R1 = R2 = R3 = R4 = 100 kΩ

2. C1 = 100 pF, CL = 20 pF

3. V+ = 5 V, V– = GND

4. CSTRAY (not shown) from Va to GND = 10 pF

Figure 39. Square Wave Oscillator Output Waveform